Mg + H2SO4 → MgSO4 + H2↑
\(n_{H_2}tt=\frac{13,44}{22,4}=0,6\left(mol\right)\)
\(\Rightarrow n_{H_2}lt=\frac{0,6}{80\%}=0,75\left(mol\right)\)
Theo pT: \(n_{Mg}=n_{H_2}lt=0,75\left(mol\right)\)
\(\Rightarrow x=m_{Mg}=0,75\times24=18\left(g\right)\)